Ratio test

Examples

Convergent because L < 1

Consider the series

${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{n}{e^n}}$

Applying the ratio test, one computes the limit

${\displaystyle L=\underset{n\to \mathrm{\infty }}{lim}\left|\frac{a_{n+1}}{a_n}\right|=\underset{n\to \mathrm{\infty }}{lim}\left|\frac{\frac{n+1}{e^{n+1}}}{\frac{n}{e^n}}\right|=\frac{1}{e}<1.}$

Since this limit is less than 1, the series converges.

Divergent because L > 1

Consider the series

${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{e^n}{n}.}$

Putting this into the ratio test:

${\displaystyle L=\underset{n\to \mathrm{\infty }}{lim}\left|\frac{a_{n+1}}{a_n}\right|=\underset{n\to \mathrm{\infty }}{lim}\left|\frac{\frac{e^{n+1}}{n+1}}{\frac{e^n}{n}}\right|=e>1.}$

Thus the series diverges.

Inconclusive because L = 1

Consider the three series

${\displaystyle \sum _{n=1}^{\mathrm{\infty }}1,}$

${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2},}$

${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^{n+1}}{n}.}$

The first series (1 + 1 + 1 + 1 + ⋯) diverges, the second (the one central to the Basel problem) converges absolutely and the third (the alternating harmonic series) converges conditionally. However, the term-by-term magnitude ratios ${\displaystyle \left|\frac{a_{n+1}}{a_n}\right|}$ of the three series are ${\displaystyle 1,}$ ${\displaystyle \frac{n^2}{(n+1{)}^2}}$ and ${\displaystyle \frac{n}{n+1}}$. So, in all three, the limit ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\left|\frac{a_{n+1}}{a_n}\right|}$ is equal to 1. This illustrates that when L = 1, the series may converge or diverge: the ratio test is inconclusive. In such cases, more refined tests are required to determine convergence or divergence.

Proof

Below is a proof of the validity of the generalized ratio test.

Suppose that ${\displaystyle r=\underset{n\to \mathrm{\infty }}{lim inf}\left|\frac{a_{n+1}}{a_n}\right|>1}$. We also suppose that ${\displaystyle (a_n)}$ has infinite non-zero members, otherwise the series is just a finite sum hence it converges. Then there exists some ${\displaystyle \ell \in (1;r)}$ such that there exists a natural number ${\displaystyle n_0\ge 2}$ satisfying ${\displaystyle a_{n_0}\ne 0}$ and ${\displaystyle \left|\frac{a_{n+1}}{a_n}\right|>\ell }$ for all ${\displaystyle n\ge n_0}$, because if no such ${\displaystyle \ell }$ exists then there exists arbitrarily large ${\displaystyle n}$ satisfying ${\displaystyle \left|\frac{a_{n+1}}{a_n}\right|<\ell }$ for every ${\displaystyle \ell \in (1;r)}$, then we can find a subsequence ${\displaystyle {\left(a_{n_k}\right)}_{k=1}^{\mathrm{\infty }}}$ satisfying ${\displaystyle \underset{n\to \mathrm{\infty }}{lim sup}\left|\frac{a_{n_k+1}}{a_{n_k}}\right|\le \ell <r}$, but this contradicts the fact that ${\displaystyle r}$ is the limit inferior of ${\displaystyle \left|\frac{a_{n+1}}{a_n}\right|}$ as ${\displaystyle n\to \mathrm{\infty }}$, implying the existence of ${\displaystyle \ell }$. Then we notice that for ${\displaystyle n\ge n_0+1}$, ${\displaystyle |a_n|>\ell |a_{n-1}|>{\ell }^2|a_{n-2}|>...>{\ell }^{n-n_0}\left|a_{n_0}\right|}$. Notice that ${\displaystyle \ell >1}$ so ${\displaystyle {\ell }^n\to \mathrm{\infty }}$ as ${\displaystyle n\to \mathrm{\infty }}$ and ${\displaystyle \left|a_{n_0}\right|>0}$, this implies ${\displaystyle (a_n)}$ diverges so the series ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}{a}_n}$ diverges by the n-th term test.
Now suppose ${\displaystyle R=\underset{n\to \mathrm{\infty }}{lim sup}\left|\frac{a_{n+1}}{a_n}\right|<1}$. Similar to the above case, we may find a natural number ${\displaystyle n_1}$ and a ${\displaystyle c\in (R;1)}$ such that ${\displaystyle |a_n|\le c^{n-n_1}\left|a_{n_1}\right|}$ for ${\displaystyle n\ge n_1}$. Then $${\displaystyle \sum _{n=1}^{\mathrm{\infty }}|a_n|=\sum _{k=1}^{n_1-1}|a_k|+\sum _{n=n_1}^{\mathrm{\infty }}|a_n|\le \sum _{k=1}^{n_1-1}|a_k|+\sum _{n=n_1}^{\mathrm{\infty }}{c}^{n-n_1}|a_{n_1}|=\sum _{k=1}^{n_1-1}|a_k|+\left|a_{n_1}\right|\sum _{n=0}^{\mathrm{\infty }}{c}^n.}$$ The series ${\displaystyle \sum _{n=0}^{\mathrm{\infty }}{c}^n}$ is the geometric series with common ratio ${\displaystyle c\in (0;1)}$, hence ${\displaystyle \sum _{n=0}^{\mathrm{\infty }}{c}^n=\frac{1}{1-c}}$ which is finite. The sum ${\displaystyle \sum _{k=1}^{n_1-1}|a_k|}$ is a finite sum and hence it is bounded, this implies the series ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}|a_n|}$ converges by the monotone convergence theorem and the series ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}{a}_n}$ converges by the absolute convergence test.
When the limit ${\displaystyle \left|\frac{a_{n+1}}{a_n}\right|}$ exists and equals to ${\displaystyle L}$ then ${\displaystyle r=R=L}$, this gives the original ratio test.

Extensions for L = 1

As seen in the previous example, the ratio test may be inconclusive when the limit of the ratio is 1. Extensions to the ratio test, however, sometimes allow one to deal with this case.

In all the tests below one assumes that ${\displaystyle \sum a_n}$ is a sum with positive ${\displaystyle a_n}$. These tests also may be applied to any series with a finite number of negative terms. Any such series may be written as:

${\displaystyle \sum _{n=1}^{\mathrm{\infty }}{a}_n=\sum _{n=1}^N{a}_n+\sum _{n=N+1}^{\mathrm{\infty }}{a}_n}$

where ${\displaystyle a_n}$ is the highest-indexed negative term. The first expression on the right is a partial sum which will be finite, and so the convergence of the entire series will be determined by the convergence properties of the second expression on the right, which may be re-indexed to form a series of all positive terms beginning at ${\displaystyle n=1}$.

Each test defines a test parameter (${\displaystyle {\rho }_n}$) which specifies the behavior of that parameter needed to establish convergence or divergence. For each test, a weaker form of the test exists which will instead place restrictions upon ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{\rho }_n}$.

All of the tests have regions in which they fail to describe the convergence properties of ${\displaystyle \sum a_n}$. In fact, no convergence test can fully describe the convergence properties of the series. This is because if ${\displaystyle \sum a_n}$ is convergent, a second convergent series ${\displaystyle \sum b_n}$ can be found which converges more slowly: i.e., it has the property that ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\frac{b_n}{a_n}=\mathrm{\infty }}$. Furthermore, if ${\displaystyle \sum a_n}$ is divergent, a second divergent series ${\displaystyle \sum b_n}$ can be found which diverges more slowly: i.e., it has the property that ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\frac{b_n}{a_n}=0}$. Convergence tests essentially use the comparison test on some particular family of ${\displaystyle a_n}$, and fail for sequences which converge or diverge more slowly.

De Morgan hierarchy

Augustus De Morgan proposed a hierarchy of ratio-type tests

The ratio test parameters (${\displaystyle {\rho }_n}$) below all generally involve terms of the form ${\displaystyle D_n{a}_n/a_{n+1}-D_{n+1}}$. This term may be multiplied by ${\displaystyle a_{n+1}/a_n}$ to yield ${\displaystyle D_n-D_{n+1}{a}_{n+1}/a_n}$. This term can replace the former term in the definition of the test parameters and the conclusions drawn will remain the same. Accordingly, there will be no distinction drawn between references which use one or the other form of the test parameter.

1. d'Alembert's ratio test

The first test in the De Morgan hierarchy is the ratio test as described above.

2. Raabe's test

This extension is due to Joseph Ludwig Raabe. Define:

${\displaystyle {\rho }_n\equiv n\left(\frac{a_n}{a_{n+1}}-1\right)}$

The series will:

• Converge when there exists a ${\displaystyle c>1}$ such that ${\displaystyle {\rho }_n\ge c}$ for all ${\displaystyle n>N}$.
• Diverge when ${\displaystyle {\rho }_n\le 1}$ for all ${\displaystyle n>N}$.
• Otherwise, the test is inconclusive.

For the limit version, the series will:

• Converge if ${\displaystyle \rho =\underset{n\to \mathrm{\infty }}{lim}{\rho }_n>1}$ (this includes the case ${\displaystyle \rho =\mathrm{\infty }}$)
• Diverge if ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{\rho }_n<1}$.
• If ${\displaystyle \rho =1}$, the test is inconclusive.

When the above limit does not exist, it may be possible to use limits superior and inferior. The series will:

• Converge if ${\displaystyle \underset{n\to \mathrm{\infty }}{lim inf}{\rho }_n>1}$
• Diverge if ${\displaystyle \underset{n\to \mathrm{\infty }}{lim sup}{\rho }_n<1}$
• Otherwise, the test is inconclusive.

Proof of Raabe's test

Defining ${\displaystyle {\rho }_n\equiv n\left(\frac{a_n}{a_{n+1}}-1\right)}$, we need not assume the limit exists; if ${\displaystyle lim sup{\rho }_n<1}$, then ${\displaystyle \sum a_n}$ diverges, while if ${\displaystyle lim inf{\rho }_n>1}$ the sum converges.

The proof proceeds essentially by comparison with ${\displaystyle \sum 1/n^R}$. Suppose first that ${\displaystyle lim sup{\rho }_n<1}$. Of course if ${\displaystyle lim sup{\rho }_n<0}$ then ${\displaystyle a_{n+1}\ge a_n}$ for large ${\displaystyle n}$, so the sum diverges; assume then that ${\displaystyle 0\le lim sup{\rho }_n<1}$. There exists ${\displaystyle R<1}$ such that ${\displaystyle {\rho }_n\le R}$ for all ${\displaystyle n\ge N}$, which is to say that ${\displaystyle \frac{a_n}{a_{n+1}}\le \left(1+\frac{R}{n}\right)\le e^{R/n}}$. Thus ${\displaystyle a_{n+1}\ge a_n{e}^{-R/n}}$, which implies that ${\displaystyle a_{n+1}\ge a_N{e}^{-R(1/N+\cdots +1/n)}\ge c{a}_N{e}^{-R\log (n)}=c{a}_N/n^R}$ for ${\displaystyle n\ge N}$; since ${\displaystyle R<1}$ this shows that ${\displaystyle \sum a_n}$ diverges.

The proof of the other half is entirely analogous, with most of the inequalities simply reversed. We need a preliminary inequality to use in place of the simple ${\displaystyle 1+t<e^t}$ that was used above: Fix ${\displaystyle R}$ and ${\displaystyle N}$. Note that ${\displaystyle \log \left(1+\frac{R}{n}\right)=\frac{R}{n}+O\left(\frac{1}{n^2}\right)}$. So ${\displaystyle \log \left(\left(1+\frac{R}{N}\right)\dots \left(1+\frac{R}{n}\right)\right)=R\left(\frac{1}{N}+\cdots +\frac{1}{n}\right)+O(1)=R\log (n)+O(1)}$; hence ${\displaystyle \left(1+\frac{R}{N}\right)\dots \left(1+\frac{R}{n}\right)\ge c{n}^R}$.

Suppose now that ${\displaystyle lim inf{\rho }_n>1}$. Arguing as in the first paragraph, using the inequality established in the previous paragraph, we see that there exists ${\displaystyle R>1}$ such that ${\displaystyle a_{n+1}\le c{a}_N{n}^{-R}}$ for ${\displaystyle n\ge N}$; since ${\displaystyle R>1}$ this shows that ${\displaystyle \sum a_n}$ converges.

(A faster approach to proving divergence: ${\displaystyle \frac{a_n}{a_{n+1}}\le \left(1+\frac{R}{n}\right)\le \left(1+\frac{1}{n}\right)}$. Thus ${\displaystyle a_{n}n\le a_{n+1}(n+1)}$, which implies that ${\displaystyle a_{n}n}$ is monotone increasing for ${\displaystyle n\ge N}$; since ${\displaystyle a_n>0}$, there must exists a constant ${\displaystyle ϵ>0}$ such that ${\displaystyle a_{n}n>ϵ}$ for all ${\displaystyle n\ge N}$. Therefore, ${\displaystyle a_n\ge \frac{ϵ}{n}}$ and ${\displaystyle \sum a_n}$ is diverges.)

3. Bertrand's test

This extension is due to Joseph Bertrand and Augustus De Morgan.

Defining:

${\displaystyle {\rho }_n\equiv n\ln n\left(\frac{a_n}{a_{n+1}}-1\right)-\ln n}$

Bertrand's test asserts that the series will:

• Converge when there exists a ${\displaystyle c>1}$ such that ${\displaystyle {\rho }_n\ge c}$ for all ${\displaystyle n>N}$.
• Diverge when ${\displaystyle {\rho }_n\le 1}$ for all ${\displaystyle n>N}$.
• Otherwise, the test is inconclusive.

For the limit version, the series will:

• Converge if ${\displaystyle \rho =\underset{n\to \mathrm{\infty }}{lim}{\rho }_n>1}$ (this includes the case ${\displaystyle \rho =\mathrm{\infty }}$)
• Diverge if ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{\rho }_n<1}$.
• If ${\displaystyle \rho =1}$, the test is inconclusive.

When the above limit does not exist, it may be possible to use limits superior and inferior. The series will:

• Converge if ${\displaystyle lim inf{\rho }_n>1}$
• Diverge if ${\displaystyle lim sup{\rho }_n<1}$
• Otherwise, the test is inconclusive.

4. Extended Bertrand's test

This extension probably appeared at the first time by Margaret Martin in 1941. A short proof based on Kummer's test and without technical assumptions (such as existence of the limits, for example) was provided by Vyacheslav Abramov in 2019.

Let ${\displaystyle K\ge 1}$ be an integer, and let ${\displaystyle {\ln }_{(K)}(x)}$ denote the ${\displaystyle K}$th iterate of natural logarithm, i.e. ${\displaystyle {\ln }_{(1)}(x)=\ln (x)}$ and for any ${\displaystyle 2\le k\le K}$, ${\displaystyle {\ln }_{(k)}(x)={\ln }_{(k-1)}(\ln (x))}$.

Suppose that the ratio ${\displaystyle a_n/a_{n+1}}$, when ${\displaystyle n}$ is large, can be presented in the form

${\displaystyle \frac{a_n}{a_{n+1}}=1+\frac{1}{n}+\frac{1}{n}\sum _{i=1}^{K-1}\frac{1}{\prod _{k=1}^i{\ln }_{(k)}(n)}+\frac{{\rho }_n}{n\prod _{k=1}^K{\ln }_{(k)}(n)},K\ge 1.}$

(The empty sum is assumed to be 0. With ${\displaystyle K=1}$, the test reduces to Bertrand's test.)

The value ${\displaystyle {\rho }_n}$ can be presented explicitly in the form

${\displaystyle {\rho }_n=n\prod _{k=1}^K{\ln }_{(k)}(n)\left(\frac{a_n}{a_{n+1}}-1\right)-\sum _{j=1}^K\prod _{k=1}^j{\ln }_{(K-k+1)}(n).}$

Extended Bertrand's test asserts that the series

• Converge when there exists a ${\displaystyle c>1}$ such that ${\displaystyle {\rho }_n\ge c}$ for all ${\displaystyle n>N}$.
• Diverge when ${\displaystyle {\rho }_n\le 1}$ for all ${\displaystyle n>N}$.
• Otherwise, the test is inconclusive.

For the limit version, the series

• Converge if ${\displaystyle \rho =\underset{n\to \mathrm{\infty }}{lim}{\rho }_n>1}$ (this includes the case ${\displaystyle \rho =\mathrm{\infty }}$)
• Diverge if ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{\rho }_n<1}$.
• If ${\displaystyle \rho =1}$, the test is inconclusive.

When the above limit does not exist, it may be possible to use limits superior and inferior. The series

• Converge if ${\displaystyle lim inf{\rho }_n>1}$
• Diverge if ${\displaystyle lim sup{\rho }_n<1}$
• Otherwise, the test is inconclusive.

For applications of Extended Bertrand's test see birth–death process.

5. Gauss's test

This extension is due to Carl Friedrich Gauss.

Assuming ${\displaystyle a_n>0}$ and ${\displaystyle r>1}$, if a bounded sequence ${\displaystyle C_n}$ can be found such that for all ${\displaystyle n}$:

${\displaystyle \frac{a_n}{a_{n+1}}=1+\frac{\rho }{n}+\frac{C_n}{n^r}}$

then the series will:

• Converge if ${\displaystyle \rho >1}$
• Diverge if ${\displaystyle \rho \le 1}$

6. Kummer's test

This extension is due to Ernst Kummer.

Let ${\displaystyle {\zeta }_n}$ be an auxiliary sequence of positive constants. Define

${\displaystyle {\rho }_n\equiv \left({\zeta }_n\frac{a_n}{a_{n+1}}-{\zeta }_{n+1}\right)}$

Kummer's test states that the series will:

• Converge if there exists a ${\displaystyle c>0}$ such that ${\displaystyle {\rho }_n\ge c}$ for all ${\displaystyle n>N}$. (Note this is not the same as saying ${\displaystyle {\rho }_n>0}$)
• Diverge if ${\displaystyle {\rho }_n\le 0}$ for all ${\displaystyle n>N}$ and ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}1/{\zeta }_n}$ diverges.

For the limit version, the series will:

• Converge if ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{\rho }_n>0}$ (this includes the case ${\displaystyle \rho =\mathrm{\infty }}$)
• Diverge if ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{\rho }_n<0}$ and ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}1/{\zeta }_n}$ diverges.
• Otherwise the test is inconclusive

When the above limit does not exist, it may be possible to use limits superior and inferior. The series will

• Converge if ${\displaystyle \underset{n\to \mathrm{\infty }}{lim inf}{\rho }_n>0}$
• Diverge if ${\displaystyle \underset{n\to \mathrm{\infty }}{lim sup}{\rho }_n<0}$ and ${\displaystyle \sum 1/{\zeta }_n}$ diverges.

Special cases

All of the tests in De Morgan's hierarchy except Gauss's test can easily be seen as special cases of Kummer's test:

• For the ratio test, let ${\displaystyle {\zeta }_n=1}$. Then:

${\displaystyle {\rho }_{\text{Kummer}}=\left(\frac{a_n}{a_{n+1}}-1\right)=1/{\rho }_{\text{Ratio}}-1}$

• For Raabe's test, let ${\displaystyle {\zeta }_n=n}$. Then:

${\displaystyle {\rho }_{\text{Kummer}}=\left(n\frac{a_n}{a_{n+1}}-(n+1)\right)={\rho }_{\text{Raabe}}-1}$

• For Bertrand's test, let ${\displaystyle {\zeta }_n=n\ln (n)}$. Then:

${\displaystyle {\rho }_{\text{Kummer}}=n\ln (n)\left(\frac{a_n}{a_{n+1}}\right)-(n+1)\ln (n+1)}$

Using ${\displaystyle \ln (n+1)=\ln (n)+\ln (1+1/n)}$ and approximating ${\displaystyle \ln (1+1/n)\to 1/n}$ for large n, which is negligible compared to the other terms, ${\displaystyle {\rho }_{\text{Kummer}}}$ may be written:

${\displaystyle {\rho }_{\text{Kummer}}=n\ln (n)\left(\frac{a_n}{a_{n+1}}-1\right)-\ln (n)-1={\rho }_{\text{Bertrand}}-1}$

• For Extended Bertrand's test, let ${\displaystyle {\zeta }_n=n\prod _{k=1}^K{\ln }_{(k)}(n)}$. From the Taylor series expansion for large ${\displaystyle n}$ we arrive at the approximation

${\displaystyle {\ln }_{(k)}(n+1)={\ln }_{(k)}(n)+\frac{1}{n\prod _{j=1}^{k-1}{\ln }_{(j)}(n)}+O\left(\frac{1}{n^2}\right)}$,

where the empty product is assumed to be 1. Then,

${\displaystyle {\rho }_{\text{Kummer}}=n\prod _{k=1}^K{\ln }_{(k)}(n)\frac{a_n}{a_{n+1}}-(n+1)\left[\prod _{k=1}^K\left({\ln }_{(k)}(n)+\frac{1}{n\prod _{j=1}^{k-1}{\ln }_{(j)}(n)}\right)\right]+o(1)=n\prod _{k=1}^K{\ln }_{(k)}(n)\left(\frac{a_n}{a_{n+1}}-1\right)-\sum _{j=1}^K\prod _{k=1}^j{\ln }_{(K-k+1)}(n)-1+o(1)}$.

Hence,

${\displaystyle {\rho }_{\text{Kummer}}={\rho }_{\text{Extended Bertrand}}-1}$.

Note that for these four tests, the higher they are in the De Morgan hierarchy, the more slowly the ${\displaystyle 1/{\zeta }_n}$ series diverges.

Proof of Kummer's test

If ${\displaystyle {\rho }_n>0}$ then fix a positive number ${\displaystyle 0<\delta <{\rho }_n}$. There exists a natural number ${\displaystyle N}$ such that for every ${\displaystyle n>N}$,

${\displaystyle \delta \le {\zeta }_n\frac{a_n}{a_{n+1}}-{\zeta }_{n+1}}$.

Since ${\displaystyle a_{n+1}>0}$, for every ${\displaystyle n>N}$,

${\displaystyle 0\le \delta a_{n+1}\le {\zeta }_n{a}_n-{\zeta }_{n+1}{a}_{n+1}}$.

In particular ${\displaystyle {\zeta }_{n+1}{a}_{n+1}\le {\zeta }_n{a}_n}$ for all ${\displaystyle n\ge N}$ which means that starting from the index ${\displaystyle N}$ the sequence ${\displaystyle {\zeta }_n{a}_n>0}$ is monotonically decreasing and positive which in particular implies that it is bounded below by 0. Therefore, the limit

${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{\zeta }_n{a}_n=L}$ exists.

This implies that the positive telescoping series

${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\left({\zeta }_n{a}_n-{\zeta }_{n+1}{a}_{n+1}\right)}$ is convergent,

and since for all ${\displaystyle n>N}$,

${\displaystyle \delta a_{n+1}\le {\zeta }_n{a}_n-{\zeta }_{n+1}{a}_{n+1}}$

by the direct comparison test for positive series, the series ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\delta a_{n+1}}$ is convergent.

On the other hand, if ${\displaystyle \rho <0}$, then there is an N such that ${\displaystyle {\zeta }_n{a}_n}$ is increasing for ${\displaystyle n>N}$. In particular, there exists an ${\displaystyle ϵ>0}$ for which ${\displaystyle {\zeta }_n{a}_n>ϵ}$ for all ${\displaystyle n>N}$, and so ${\displaystyle \sum _n{a}_n=\sum _n\frac{a_n{\zeta }_n}{{\zeta }_n}}$ diverges by comparison with ${\displaystyle \sum _n\frac{ϵ}{{\zeta }_n}}$.

Tong's modification of Kummer's test

A new version of Kummer's test was established by Tong. See also for further discussions and new proofs. The provided modification of Kummer's theorem characterizes all positive series, and the convergence or divergence can be formulated in the form of two necessary and sufficient conditions, one for convergence and another for divergence.

• Series ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}{a}_n}$ converges if and only if there exists a positive sequence ${\displaystyle {\zeta }_n}$, ${\displaystyle n=1,2,\dots }$, such that ${\displaystyle {\zeta }_n\frac{a_n}{a_{n+1}}-{\zeta }_{n+1}\ge c>0.}$

• Series ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}{a}_n}$ diverges if and only if there exists a positive sequence ${\displaystyle {\zeta }_n}$, ${\displaystyle n=1,2,\dots }$, such that ${\displaystyle {\zeta }_n\frac{a_n}{a_{n+1}}-{\zeta }_{n+1}\le 0,}$ and ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{1}{{\zeta }_n}=\mathrm{\infty }.}$

The first of these statements can be simplified as follows:

• Series ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}{a}_n}$ converges if and only if there exists a positive sequence ${\displaystyle {\zeta }_n}$, ${\displaystyle n=1,2,\dots }$, such that ${\displaystyle {\zeta }_n\frac{a_n}{a_{n+1}}-{\zeta }_{n+1}=1.}$

The second statement can be simplified similarly:

• Series ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}{a}_n}$ diverges if and only if there exists a positive sequence ${\displaystyle {\zeta }_n}$, ${\displaystyle n=1,2,\dots }$, such that ${\displaystyle {\zeta }_n\frac{a_n}{a_{n+1}}-{\zeta }_{n+1}=0,}$ and ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{1}{{\zeta }_n}=\mathrm{\infty }.}$

However, it becomes useless, since the condition ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{1}{{\zeta }_n}=\mathrm{\infty }}$ in this case reduces to the original claim ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}{a}_n=\mathrm{\infty }.}$

Frink's ratio test

Another ratio test that can be set in the framework of Kummer's theorem was presented by Orrin Frink 1948.

Suppose ${\displaystyle a_n}$ is a sequence in ${\displaystyle \mathbb{C}\setminus \{0\}}$,

• If ${\displaystyle \underset{n\to \mathrm{\infty }}{lim sup}(\frac{|a_{n+1}|}{|a_n|}{)}^n<\frac{1}{e}}$, then the series ${\displaystyle \sum _n{a}_n}$ converges absolutely.
• If there is ${\displaystyle N\in \mathbb{N}}$ such that ${\displaystyle (\frac{|a_{n+1}|}{|a_n|}{)}^n\ge \frac{1}{e}}$ for all ${\displaystyle n\ge N}$, then ${\displaystyle \sum _n|a_n|}$ diverges.

This result reduces to a comparison of ${\displaystyle \sum _n|a_n|}$ with a power series ${\displaystyle \sum _n{n}^{-p}}$, and can be seen to be related to Raabe's test.

Ali's second ratio test

A more refined ratio test is the second ratio test: For ${\displaystyle a_n>0}$ define:

By the second ratio test, the series will:

• Converge if ${\displaystyle L<\frac{1}{2}}$
• Diverge if ${\displaystyle L>\frac{1}{2}}$
• If ${\displaystyle L=\frac{1}{2}}$ then the test is inconclusive.

If the above limits do not exist, it may be possible to use the limits superior and inferior. Define:

Then the series will:

• Converge if ${\displaystyle L<\frac{1}{2}}$
• Diverge if ${\displaystyle \ell >\frac{1}{2}}$
• If ${\displaystyle \ell \le \frac{1}{2}\le L}$ then the test is inconclusive.

Ali's mth ratio test

This test is a direct extension of the second ratio test. For ${\displaystyle 0\le k\le m-1,}$ and positive ${\displaystyle a_n}$ define:

By the ${\displaystyle m}$th ratio test, the series will:

• Converge if ${\displaystyle L<\frac{1}{m}}$
• Diverge if ${\displaystyle L>\frac{1}{m}}$
• If ${\displaystyle L=\frac{1}{m}}$ then the test is inconclusive.

If the above limits do not exist, it may be possible to use the limits superior and inferior. For ${\displaystyle 0\le k\le m-1}$ define:

Then the series will:

• Converge if ${\displaystyle L<\frac{1}{m}}$
• Diverge if ${\displaystyle \ell >\frac{1}{m}}$
• If ${\displaystyle \ell \le \frac{1}{m}\le L}$, then the test is inconclusive.

Ali--Deutsche Cohen φ-ratio test

This test is an extension of the ${\displaystyle m}$th ratio test.

Assume that the sequence ${\displaystyle a_n}$ is a positive decreasing sequence.

Let ${\displaystyle \phi :{\mathbb{Z}}^{+}\to {\mathbb{Z}}^{+}}$ be such that ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\frac{n}{\phi (n)}}$ exists. Denote ${\displaystyle \alpha =\underset{n\to \mathrm{\infty }}{lim}\frac{n}{\phi (n)}}$, and assume ${\displaystyle 0<\alpha <1}$.

Assume also that ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\frac{a_{\phi (n)}}{a_n}=L.}$

Then the series will:

• Converge if ${\displaystyle L<\alpha }$
• Diverge if ${\displaystyle L>\alpha }$
• If ${\displaystyle L=\alpha }$, then the test is inconclusive.